1000 Solved Problems in Modern Physics

1.3 Solutions 61

Fig. 1.16Two modes of Oscillation

If initiallyx 1 =x 2 , the masses oscillate in phase with frequencyω 1 (sym-

metrical mode) as in Fig. 1.16(a). If initiallyx 2 =−x 2 then the masses

oscillate out of phase (asymmetrical) as in Fig. 1.16(b)

1.62 Sum of translational+rotational+potential energy=constant
1
2

mv^2 +

1

2

Iω^2 +

1

2

kx^2 =const.

ButI=

1

2

mR^2 andω=v/R

Therefore

3

4

mv^2 +

1

2

kx^2 =const.

3

4

m(dx/dt)^2 +

1

2

kx^2 =const.

Differentiating with respect to time,

(

3

2

)(

md^2 x

dt^2

)

.

dx

dt

+kx.

dx

dt

= 0

Cancelling dx/dtthrough and simplifying d^2 x/dt^2 +(2k/ 3 m)x=0. This

is an equation to SHM.

Writingω^2 = 32 mk, time periodT=^2 ωπ= 2 π

√

3 m

2 k

1.63

d^2 y

dx^2

− 8

dy

dx

=− 16 y

d^2 y

dx^2

− 8

dy

dx

+ 16 y= 0

Auxiliary equation:

D^2 − 8 D+ 16 = 0

(D−4)(D−4)= 0

The roots are 4 and 4.

Thereforey=C 1 e^4 x+C 2 xe^4 x

1.64x^2

dy

dx

+y(x+1)x= 9 x^2 (1)

Put the above equation in the form